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JEE Advance 2021
Paper-2
Question
A special metal S conducts electricity without any resistance. A closed wire loop, made of S , does not allow any change in flux through itself by inducing a suitable current to generate a compensating flux. The induced current in the loop cannot decay due to its zero resistance. This current gives rise to a magnetic moment which in turn repels the source of magnetic field or flux. Consider such a loop, of radius $a$, with its center at the origin. A magnetic dipole of moment $m$ is brought along the axis of this loop from infinity to a point at distance $\mathrm{r}(\gg \mathrm{a})$ from the center of the loop with its north pole always facing the loop, as shown in the figure below. The magnitude of magnetic field of a dipole m , at a point on its axis at distance $\mathrm{r}_1$ is $\frac{\mu_0}{2 \pi} \frac{\mathrm{~m}}{\mathrm{r}^3}$, where $\mu_0$ is the permeability of free space. The magnitude of the force between two magnetic dipoles with moments, $m_1$ and $m_2$, separated by a distance $r$ on the common axis, with their north poles facing each other, is $\frac{k m_1 m_2}{r^4}$, where $k$ is a constant of appropriate dimensions. The direction of this force is along the line joining the two dipoles.

The work done in bringing the dipole from infinity to a distance $r$ from the center of the loop by the given process is proportional to
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Select the correct option:
A
$\frac{m}{r^5}$
B
$\frac{m^2}{r^5}$
C
$\frac{m^2}{r^6}$
D
$\frac{m^2}{r^7}$
✓ Correct! Well done.
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Solution
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Question Tags
JEE Advance
Physics
Easy
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